# Can you tell how far a photon has traveled?

1. Dec 19, 2004

### ktpr2

is it possible to determine how far a photon has travelled if you do not know where it "started" from? I'm thinking if photons decay at a consistent rate then it might be possible.

2. Dec 19, 2004

### mathman

Since photons can start at any frequency (also for the value of other properties), there is no way to tell how far a photon has travelled just from its detection. By looking back along its path until you see something you may be able to find its origin.

3. Jan 4, 2005

### IKnowGravity

Photons and frekvensy

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They can go along way, forever without diminishing in intensity. Only if they hit something dead on do they spend their energy. They can however push matter before it if they are close to it without diminishing in energy as their energy is in their frekvency. And their frekvensy is not diminishing with distanse. Their elektrik and magnetik componant is what affekts matter. :!!)

4. Jan 7, 2005

### LURCH

I would only add that photons are not known to deteriorate or lose energy as they travel. There are some theories that postulate that photons lose energy and experience an increase of wavelength (decreased frequency) over great amounts of time/distance. These predictions are, by their very nature extremely difficult to support or refute experimentally, and are not generally accepted.

5. Jan 7, 2005

### DB

Unless you are talking about photons in a vaccum, then photons do lose their energy (frequency), (increase in wavelenght) as they travel through space and therefore through gravitational fields. Since photons can't be stoped or slowed, they lose energy to remain propagating at c. Though they usual lose minimal amounts of energy, they are still redshifted.

6. Jan 8, 2005

### Mk

What about working backwards to find interference patterns, then to where the particle orginated?

7. Jan 18, 2005

### nightcleaner

I had a slight idea along these lines but it probably will only reveal my weak understanding of these matters. Anyway here it is.

Photons from very far away are also from very long ago. The universe has expanded a great deal since the far away photons were created. We might imagine that the photons from furthest away and longest ago have been expanded along with the universe, so that they would now be of very low energy. So we would have to look for photons with the longest possible wavelength. Maybe we would look for photons above the radio broadcast band, more than several hundred meters long.

Then, if we could detect such photons in sufficient numbers from a single source, we might be able to look at the emission spectra. Now what I am thinking is, and here is where I suspect I may have gone wrong, that the emission spectra, in addition to being redshifted, will also show a widening of the spectral lines, reflecting the expansion of space since they were created. Then if we knew the history of the expansion of space (inflation) we could get an idea of how far away, and how long ago, the photons were created.

I would be interested to hear if anyone thinks I could be right, or if someone knows why this expectation is in error.

Thanks,

nc

8. Jan 18, 2005

### chroot

Staff Emeritus
nightcleaner,

The earliest photons in the universe that have survived to the present are those that now compose the cosmic microwave background radiation (CMBR), predominantly in the microwave part of the electromagnetic spectrum today. The problem is that those photons were thermal photons, emitted haphazardly by a very hot plasma about 300,000 years after the big bang. There are no emission lines because the dominant radiators were not atoms, but electrons undergoing free-free emission. These CMBR photons existed before the first electrons fell into place around nuclei. Only atoms, with their discretized electronic energy levels, can produce discretized spectral lines -- free electrons produce continuum radiation, with no distinguishing spectral features.

- Warren

9. Jan 18, 2005

### Chronos

I would give a qualified 'yes' to the original question. Light from remote sources is redshifted due to expansion of the universe. Since we have measured the rate of expansion [~72 km/Mpc], you can tell how far the photon has traveled since it was emitted by its redshift - if, as chroot noted, you can find an emission line.

'Tired light' is an old, vagrant [no visible means of support] hypothesis. It is widely regarded as incorrect. Energy cannot be lost without something else gaining the same amount of energy. Only mass possessing entities participate in energy exchanges. Energy ultimately comes down to making something move - such as an electron.

Gravitational braking [gravitational redshift] of photons only occurs as a photon in emitted. Once on its way, the only effect gravity of objects along its path has is lensing. The frequency does not change. As a photon approaches a gravitational field, it is blueshifted. As it exits the gravity field, it is redshifted back to its original frequency.

Last edited: Jan 18, 2005
10. Jan 19, 2005

### nightcleaner

I am still confused. If the light is red shifted into longer wavelengths, is it not also shifted to a lower energy? Wouldn't the light emitted by free electrons be very high energy, say in the x-ray part of the spectrum? Near the beginning of the universe, weren't the particles very densely packed, and so the wavelengths were very short, that is very high energy? And now they arrive to us as microwaves, very much red shifted from their original energy? This red shift and the loss of energy are due to the expansion of the spacetime fabric?

And as for emission lines, if we can detect the CMBE, should we not also be able to detect emission lines from atoms slightly later in cosmic history? In fact, if all those free electrons were falling suddenly into orbitals, would they not release really a lot of photons as they dropped through the orbital levels? Is an electron dropping into an orbital sufficient to release a photon, or does it have to be an electron excited out of an orbital and then dropping back in that creates the photon? Is there a difference between a photon created by dropping in and a photon created by getting kicked out?

To recap,

1. Did the cmbe start out as high energy xrays?
2. If so, what happened to the energy as they were redshifted to the present microwaves?
3. Could we suppose that the lost energy went into the work of expanding the universe?
4. If we can detect the cmbe, do we also detect photons due to later collapse of the electrons into orbitals? The only candidates that come to my innocent mind are quasars. Then why do we detect quasars as points while we detect cmbe as a universally dispersed themal energy?
5. Is it possible that there is really only one quasar and we see it in many different directions because the light from the quasar has gone all the way around the universe? Then the geometric distribution of quasars would tell us something about the surface conditions of the universe....light from the electron collapse arrives preferentially from the directions where we see the quasars?
6. Is a photon the result of an electron dropping into an orbital, or of getting excited into a higher orbital, or either, or only both together? If either, is there a difference between the photon emitted from the falling electron and the photon emitted by the rising electron? Does the rise and fall of an electron emit two photons? Does a photon emitted by a rising and/or falling electron radiate out in a single direction or does it radiate outward equally in all directions? I know statistically they radiate in all directions, but if we can talk about a single electron rising or falling, can we not also talk of a single photon being produced? And if a single photon is produced, does it have a specified direction, or does it radiate like a wave on water, in every possible direction? If we detect an incoming photon from a detector like the Very Large Array, is it detected by only one or two of the antennae or is it detected collectively by all of them together? When a photon strikes a phosphorescent screen, is it striking only that one spot or is it striking the entire screen? I am thinking of QED, where a photon is thought of as striking the entire surface of a glass sheet, not really just a single point as we imagine in calculations using Snell's law of refraction.

Well these are a lot of questions i have been wondering about for a long time. If the learned gentlefolk would comment on any of them, I would be gratified.

Thanks,

nc

11. Jan 19, 2005

### chroot

Staff Emeritus
nightcleaner,

Lots of excellent questions here! I'll do my best to answer them.

In the very early universe, nearly all the radiation was originally in the gamma part of the spectrum, since temperatures were extremely high.
That's correct.
The production of an emission line requires a "pump" -- some mechanism which continually adds energy to atoms, exciting their electrons to more energetic orbitals. If you want an emission line signal to last for millions or billions of years, you have to have some kind of a pump that operates for millions or billions of years. There was no such pump in the early universe; the atoms were certainly excited by stray gammas and thermal collisions, but there was no effective, organized pump in place.
They would -- and those photons would be members of that atom's spectrum. The problem is that, for most atoms, this coming-together (perversely called "recombination") happened only one time, releasing only one photon. The early universe was "radiation-dominated," in that the vast majority of the early universe's energy was in the form of radiation. There were billions or trillions (I can probably look up a specific number if you'd like) of thermal photons for every atom. Recombination did produce some spectral emissions, but they were totally swamped by that overwhelming random thermal radiation.

Nope, a free electron becoming bound has to lose its energy somehow -- it does so by emitting a photon.
Nope. Photons are photons -- they just come in different frequencies. There's no way to tell two photons of the same frequency (and polarization) apart.
Yes.
Cosmological redshift.
Yes, although it would be more accurate to say the energy went into the work of slowing the universe's expansion.
The signal to noise ratio is too much small to ever make such a measurement possible.
Quasars are just juvenile galaxies with active cores -- they have nothing to do with the CMBR. Quasars didn't come onto the scene until (probably) hundreds of millions of years after recombination.
Since the universe is not infinitely old, and the speed of light is not infinite, we can only see part of the universe, that within about 43 billion light-years' distance. The entire universe might be curved enough to allow light to "circumnavigate" it, but light from one quasar has certainly not had time to propagate round-trip around even around the part of the universe accessible to us.
It takes energy to get an electron into a higher orbital, but the atom will emit a photon when that photon falls back to a lower energy level.
Macroscopically, a single photon only goes in a single direction, rather like a bullet. Of course, if you make the photon contend with very small obstructions, you will begin to see its wave nature -- it will diffract, for example. On a cosmological scale, however, you can just think of photons as bullets.
A single, individual photon will be detected by only one antenna at a time.
QED deals with "virtual photons" which travel all possible paths, including those that would violate the speed of light. The statistical contributions from all these paths are summed, and the result is the path that a real photon could travel. Even if you'd like to think of photons as simultaneously taking all possible paths to "decide" where to arrive, they indisputably arrive at only one spot on your detector screen.

- Warren

Last edited: Jan 19, 2005
12. Jan 19, 2005

### ahrkron

Staff Emeritus
This last one touches the idea of the collapse of the wavefunction: before detection, the photon can be represented as a superposition of many position eigenfunctions. The process of measuring position gets rid of all but one of these components.

13. Jan 20, 2005

### nightcleaner

Thank you, I believe my idea of photons is somewhat clearer now. Perhaps I may say that what a photon "is", concurrent with the observor, is a wave, not localized in space, but what it "was", once the observation has been made, is a specific particle with a specific location.

I remember thinking, the last time I read Feynman's QED, that the size of the reflective surface is directly related to the resolution of the image even when the image is composed of a series of single photons because the larger surface area encounters more of the virtual photons, thereby making the collapse of the wavefunction more precise. This would apply by extension of the idea to the VLA, where a large collection area, even though the individual antennae are not larger, would make the resolution of the image better, where in deep field observations the photons are collected a few at a time over a very long observation period.

If I have this correctly, one could build a large mirror of small unit tiles, then collect the image of a point source of light, which of course would be seen reflected in only a single tile, and then conduct an experiment by removing a few of the tiles at a time from a region of the large mirror at a distance from the apparent image, and record any changes this made in the resolution of the image of the point source. If I have this correctly, removing tiles from the outer parts of the large mirror would cause the image of the point to become more diffuse as more tiles were removed, even though the removed tiles were not directly in the line of the image.

An increasingly diffuse image as tiles were removed would support the physical reality of virtual photons.

What if the light source were a laser? Since a laser beam does not spread out as it propogates, are virtual photons absent? I remember many years ago a physics professor at my university had obtained a laser, a great curiosity at that time, and in class he tried shining it on a prism. It was a dangerous experiment, as I now know. The professor shut the thing down almost instantly, as he must have realized the danger also, but in the short observation period I had, as I recall, the sensation of points of laser light coming at me from many directions at once, almost as if the prism had exploded the wavefunction of the light rather than collapsing it. But it was just an excited instant and memory does not always serve the cause of science.

Thanks again for the detailed exposition.

nc

14. Jan 20, 2005

### chroot

Staff Emeritus
There's no need to do a specific experiment -- we already know quite well that small dishes separated by a large distance -- for example, two small dishes 5,000 miles apart -- achieve the same resolution as a theoretical dish 5,000 miles wide.

Laser light actually does spread out over distance -- just not as much as, say, a flashlight.

A prism won't do anything to laser light at all but change its ultimate direction; laser light is composed of only one frequency (color) and thus will not be spread out by the prism. It sounds to me like your professor's prism was being moved around, and the beam was bouncing all over the room -- bad idea.

- Warren

15. Jan 21, 2005

### nightcleaner

It was a large auditoreum and as I recall the beam and the prism both were held by hand.

I guess I was thinking of how a student might build a science fair project to try to demonstrate the virtual particles. It would be a fussy experiment, all those little tiles, and as you say would give nothing new. Oh, I guess you wouldn't have to use little tiles, you could just use a large mirror and blacken out the outer edges. Now that I think of it, lenses work according to a similar principle in terms of virtual photons. You could set up a desk top apparatus using an iris. Come to think of it, my microscope has an iris. Turns out there is a way to open and close the iris over the lens to restrict the light in such a way that the image is sharpest. A litte more open, less resolution, a little more closed, less resolution, but at just the right setting, sharpest image.

what happens if you shine a laser through a microscope? I know, I know, bad idea, you could put your eye out with that thing. Always fun until somebody gets hurt. Funny as a poke in the eye with a sharp stick. Oh well. Maybe that's why I don't teach science.

thanks,

nc